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$\begin{array}{l} A mass m= 0 .55 kg rests on a horizontal surface . The coefficient of static friction between \\ the mass and the surface is 0 .4, if the mass is pulled by a force F= \sqrt{3} N as shown in the figure . \\ The limiting friction between mass and the surface will be (g = 10 m / s^{2}) . \end{array}$

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1.6 N

16 N

$\frac{\sqrt{3}}{2}$ N

0 N

answer is A.

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Detailed Solution

$\begin{array}{l} N = mg - Fsin 60^{0} = 0 .55 \times 10 - \sqrt{3} \times \frac{\sqrt{3}}{2} = 4 \\ {Limiting friction f}_{l} = μ_{s} N = 0 .4 \times 4 = 1 .6 newton \end{array}$

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